apoorv
Posts:
89
Registered:
6/18/05


Is the set N of natural numbers well defined?
Posted:
Jun 18, 2005 3:08 AM


The set N is defined as the set of all finite ordinals.Every ordinal is the set of all preceding ordinals. In that case would not N contain the infinite ordinal N1? And thereby contradict its own definition as the set of all finite ordinals? The conceptual number system is set up in steps, starting from unity. The axiom of infinity asserts that at every stage, we can take the next step. The set N is defined as the set of all finite ordinals. When we speak of the set N, we are asserting that all the steps have been taken.Is there is a logical weakness here? if we can always take the Next Step, can we reach a stage where we have taken All the Steps? Once we accept N, we are still faced with the problem of the next step; we then create a hierarchy of infinities. The problem of the next step is still not wished away and is contained in the BuraliForti paradox. The specification of N as the set of all finite ordinals leads to some rather paradoxical results. The foremost of these is the result of Godel and Cohen on the Continuum Hypothesis that the existence and non existence of an uncountable subset of R with cardinality less than c are both consistent with the axioms of set theory. If, by the axioms of extent and specification, a set is completely well defined by its elements and hence its subsets, than does this mean that R and hence N is not well defined? We consider the sets R defined as R1 ={ },R2 ={{ }},and so on, and the sets S defined as S ={R1 },S ={R1 ,R2}, and so on.
Then the following are true statements: 1. If Ri does not belong to Ri , then, R(i+1) does not belong to R(i+1) 2. Ri belongs to Si , but R(i+1) does not belong to Si . 3. There is a 1 to 1 correspondence between the sets R and the sets S .
Let us consider the set Q of all the R that do not contain themselves. Then the set Q is not empty. It is also one of the Ss say Sa . Then, there exists a Ra that corresponds to Sa. Now, Ra belongs to Sa but R(a+1) does not belong to Sa ,(by 2 above) Ra does not belong to Ra (by definition of S ) Therefore, R(a+1) does not belong to R(a+1) (by 1 above). Therefore, R(a+1) belongs to Sa . Thus we have a contradiction!
Thus the set S is not well defined. But S is a representation of the set N of all finite numbers. Thus this representation of the set N is not well defined.
The set S is precisely the set constructed by Bertrand Russell. However, the fact that Q (or Sa) is not well defined was used to argue against the existence of a set containing all sets rather than against the existence of a set of all finite sets. Alternately, consider the class of all ordinals. We note that some ordinals do not contain themselves; however, the ordinal w contains w1 and hence itself. We consider the set T of all ordinals that do not contain themselves. Then, T is itself an ordinal. Now T belongs to T, if T does not belong to T and vice versa. Thus, T is not well defined. But T is precisely the set N. The above arguments can be more informally put as follows: The conceptual universe is created in steps from unity. If we move a step forward from the finite, we are still in the finite. If we move backward a step from infinity, we are still in the infinite. Can we consistently speak of the All of the Finite? If All of the Finite is finite, then the next step is still in the finite, and All of the Finite does not contain some of the finite. If All of the Finite is infinite, then a step backward still keeps us in the infinite, and All of the Finite contains the infinite as well. Thus, can we consistently speak of the Allof the Finite?

